Linear Algebra: Exercise 6 - Proving Equivalence of Systems

[carlosbgois]

This is a question from 'Linear Algebra' by Hoffman and Kunze, and it's exercise 6 at page 5:

Question: Prove that if two homogeneous systems of linear equations in two unknowns
have the same solutions, then they are equivalent.

Attempt: I've tried some approaches but I couldn't even start.

Does anyone have a clue? Is there any solutions manual for this book?
Thanks

 

[micromass]

Hi carlosbgois!

This is a question from 'Linear Algebra' by Hoffman and Kunze, and it's exercise 6 at page 5:

Question: Prove that if two homogeneous systems of linear equations in two unknowns
have the same solutions, then they are equivalent.

Attempt: I've tried some approaches but I couldn't even start.

Does anyone have a clue? Is there any solutions manual for this book?
Thanks

The solution is a bit complicated, but let's guide you through it.

Firstly, two systems of equations are equivalent if each row of the system is a linear combination of the rows of the other system.

So take two systems

$$\left\{\begin{array}{c} ax+by=0\\ cx+dy=0\end{array}\right.~\text{and}~\left\{\begin{array}{c} a^\prime x+b^\prime y=0\\ c^\prime x+d^\prime y=0\end{array}\right.$$

So you must prove that there exists $\alpha,\beta$ such that

$$ax+by=\alpha (a^\prime x+b^\prime y)+\beta (c^\prime x+d^\prime)$$

This leads us to the following system of equations

$$\left\{ \begin{array}{c} \alpha a^\prime+\beta c^\prime = a\\ \alpha b^\prime+\beta d^\prime = b \end{array} \right.$$

And we must prove that this system has a solution. So, how do we do that? Well, by solving the system!

 

[carlosbgois]

Thanks for helping, but I didn't quite grasp it yet.

You started supposing that the two systems are equivalent, and then managed to show that they have the same solutions, right? But is the reciprocal also true? Because I need to prove that if they have the same answer they are equivalent, not the inverse as done.

And I also didn't get how

$$ax+by=\alpha (a^\prime x+b^\prime y)+\beta (c^\prime x+d^\prime)$$

Leads us to

$$\left\{ \begin{array}{c} \alpha a^\prime+\beta c^\prime = a\\ \alpha b^\prime+\beta d^\prime = b \end{array} \right.$$

 

[micromass]

Thanks for helping, but I didn't quite grasp it yet.

You started supposing that the two systems are equivalent, and then managed to show that they have the same solutions, right? But is the reciprocal also true? Because I need to prove that if they have the same answer they are equivalent, not the inverse as done.

What I'm trying to find out first is a criterion of when the two systems are equivalent. So far, we have shown that the two systems are equivalent if and only if

$$\left\{ \begin{array}{c} \alpha a^\prime+\beta c^\prime = a\\ \alpha b^\prime+\beta d^\prime = b \end{array} \right.$$

has a solution. So what I ask from you is to find out what this solution is and when it exists.

And I also didn't get how

$$ax+by=\alpha (a^\prime x+b^\prime y)+\beta (c^\prime x+d^\prime)$$

Leads us to

$$\left\{ \begin{array}{c} \alpha a^\prime+\beta c^\prime = a\\ \alpha b^\prime+\beta d^\prime = b \end{array} \right.$$

Group all the x-terms and all the y-terms. The equality of the coefficient of the x-terms and of the y-terms is this system. Thus, if we work out

$$ax+by=\alpha (a^\prime x+b^\prime y)+\beta (c^\prime x+d^\prime),$$

then we get

$$ax+by=(\alpha a^\prime+\beta c^\prime) x+(\alpha b^\prime +\beta d^\prime )$$

So since the coefficients of the x-terms must equal each other, we get that

$$a=\alpha a^\prime+\beta c^\prime$$

Analogous with the coefficients of the y-terms.